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MCK's questions at Yahoo! Answers regarding an inexact ODE

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MarkFL

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Feb 24, 2012
13,775
Here is the question:

Find the general solution of the following differential equation...?

y+3y^7=(y^5+6x)y′.

Write your solution in the form F(x,y)=C, where C is an arbitrary constant.

Thank you.
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
13,775
Hello MCK,

I would first write the ODE in differential form:

\(\displaystyle \left(y+3y^7 \right)dx+\left(-y^5-6x \right)dy=0\)

Next, let us compute:

\(\displaystyle \frac{\partial}{\partial y}\left(y+3y^7 \right)=1+21y^6\)

\(\displaystyle \frac{\partial}{\partial x}\left(-y^5-6x \right)=-6\)

Hence, we find the equation is inexact. So we look for an integrating factor. To do this, we consider:

\(\displaystyle \frac{(-6)-\left(1+21y^6 \right)}{y+3y^7}=-\frac{21y^6+7}{3y^7+y}\)

Since this is a function of just $y$, we find our special integrating factor with:

\(\displaystyle \mu(y)=\exp\left(\int -\frac{21y^6+7}{3y^7+y}\,dy \right)\)

To compute the integrating factor, we need to compute:

\(\displaystyle \int \frac{21y^6+7}{3y^7+y}\,dy\)

Let's write the integrand as:

\(\displaystyle \frac{21y^6+7}{3y^7+y}=\frac{21y^6+1}{3y^7+y}+ \frac{6}{3y^7+y}\)

The first term is easily integrated, and for the second term, let's use partial fraction decomposition to write:

\(\displaystyle \frac{6}{3y^7+y}=\frac{6}{y}-\frac{18y^5}{3y^6+1}\)

And so, putting all of this together, we find:

\(\displaystyle \mu(y)=e^{-\left(\ln\left|3y^7+y \right|+6\ln|y|-\ln\left|3y^6+1 \right| \right)}\)

Hence:

\(\displaystyle \mu(y)=\frac{3y^6+1}{y^6\left(3y^7+y \right)}=\frac{1}{y^7}\)

Multiplying the ODE by this factor, we obtain:

\(\displaystyle \left(y^{-6}+3 \right)dx+\left(-y^{-2}-6xy^{-7} \right)dy=0\)

Now, it is easy to see that we have an exact ODE. Hence:

\(\displaystyle \frac{\partial F}{\partial x}=y^{-6}+3\)

Integrating with respect to $x$, we find:

\(\displaystyle F(x,y)=xy^{-6}+3x+g(y)\)

Now, to determine $g(y)$, we differentiate with respect to $y$, recalling \(\displaystyle \frac{\partial F}{\partial x}=-y^{-2}-6xy^{-7}\):

\(\displaystyle -y^{-2}-6xy^{-7}=-6xy^{-7}+g'(y)\)

Thus, we find:

\(\displaystyle g'(y)=-y^{-2}\)

Hence, by integrating, we obtain:

\(\displaystyle g(y)=y^{-1}\)

And so we have:

\(\displaystyle F(x,y)=xy^{-6}+3x+y^{-1}\)

And so the solution to the ODE is given implicitly by:

\(\displaystyle xy^{-6}+3x+y^{-1}=C\)

edit: Hello MCK, I saw your message at Yahoo asking me to take a look at another of your questions posted there, however, Yahoo will not let me access the link, nor will it let me post a comment there. As an alternative, you could post the question here in our differential equations forum.